3.5.52 \(\int \frac {-192 x^2-64 x^3-12 e^2 x^4+e (-96 x^3-12 x^4)+e^x (48 x^2+16 x^3-x^4+3 e^2 x^4+e (24 x^3+3 x^4))+(-96 x^2-16 x^3-24 e x^3+e^x (24 x^2+4 x^3+6 e x^3)) \log (4-e^x)+(-12 x^2+3 e^x x^2) \log ^2(4-e^x)}{-64-32 e x-4 e^2 x^2+e^x (16+8 e x+e^2 x^2)+(-32-8 e x+e^x (8+2 e x)) \log (4-e^x)+(-4+e^x) \log ^2(4-e^x)} \, dx\)

Optimal. Leaf size=23 \[ x^3+\frac {x^4}{4+e x+\log \left (4-e^x\right )} \]

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Rubi [F]  time = 3.22, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-192*x^2 - 64*x^3 - 12*E^2*x^4 + E*(-96*x^3 - 12*x^4) + E^x*(48*x^2 + 16*x^3 - x^4 + 3*E^2*x^4 + E*(24*x^
3 + 3*x^4)) + (-96*x^2 - 16*x^3 - 24*E*x^3 + E^x*(24*x^2 + 4*x^3 + 6*E*x^3))*Log[4 - E^x] + (-12*x^2 + 3*E^x*x
^2)*Log[4 - E^x]^2)/(-64 - 32*E*x - 4*E^2*x^2 + E^x*(16 + 8*E*x + E^2*x^2) + (-32 - 8*E*x + E^x*(8 + 2*E*x))*L
og[4 - E^x] + (-4 + E^x)*Log[4 - E^x]^2),x]

[Out]

x^3 - (1 + E)*Defer[Int][x^4/(4 + E*x + Log[4 - E^x])^2, x] - 4*Defer[Int][x^4/((-4 + E^x)*(4 + E*x + Log[4 -
E^x])^2), x] + 4*Defer[Int][x^3/(4 + E*x + Log[4 - E^x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2 \left (12 e^2 x^2-3 e^{2+x} x^2+64 (3+x)+12 e x (8+x)-3 e^{1+x} x (8+x)-e^x \left (48+16 x-x^2\right )-2 \left (-4+e^x\right ) (12+(2+3 e) x) \log \left (4-e^x\right )-3 \left (-4+e^x\right ) \log ^2\left (4-e^x\right )\right )}{\left (4-e^x\right ) \left (4+e x+\log \left (4-e^x\right )\right )^2} \, dx\\ &=\int \left (-\frac {4 x^4}{\left (-4+e^x\right ) \left (4+e x+\log \left (4-e^x\right )\right )^2}+\frac {x^2 \left (48+16 \left (1+\frac {3 e}{2}\right ) x-(1-3 e (1+e)) x^2+24 \log \left (4-e^x\right )+4 \left (1+\frac {3 e}{2}\right ) x \log \left (4-e^x\right )+3 \log ^2\left (4-e^x\right )\right )}{\left (4+e x+\log \left (4-e^x\right )\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {x^4}{\left (-4+e^x\right ) \left (4+e x+\log \left (4-e^x\right )\right )^2} \, dx\right )+\int \frac {x^2 \left (48+16 \left (1+\frac {3 e}{2}\right ) x-(1-3 e (1+e)) x^2+24 \log \left (4-e^x\right )+4 \left (1+\frac {3 e}{2}\right ) x \log \left (4-e^x\right )+3 \log ^2\left (4-e^x\right )\right )}{\left (4+e x+\log \left (4-e^x\right )\right )^2} \, dx\\ &=-\left (4 \int \frac {x^4}{\left (-4+e^x\right ) \left (4+e x+\log \left (4-e^x\right )\right )^2} \, dx\right )+\int \left (3 x^2-\frac {(1+e) x^4}{\left (4+e x+\log \left (4-e^x\right )\right )^2}+\frac {4 x^3}{4+e x+\log \left (4-e^x\right )}\right ) \, dx\\ &=x^3-4 \int \frac {x^4}{\left (-4+e^x\right ) \left (4+e x+\log \left (4-e^x\right )\right )^2} \, dx+4 \int \frac {x^3}{4+e x+\log \left (4-e^x\right )} \, dx+(-1-e) \int \frac {x^4}{\left (4+e x+\log \left (4-e^x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 23, normalized size = 1.00 \begin {gather*} x^3+\frac {x^4}{4+e x+\log \left (4-e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-192*x^2 - 64*x^3 - 12*E^2*x^4 + E*(-96*x^3 - 12*x^4) + E^x*(48*x^2 + 16*x^3 - x^4 + 3*E^2*x^4 + E*
(24*x^3 + 3*x^4)) + (-96*x^2 - 16*x^3 - 24*E*x^3 + E^x*(24*x^2 + 4*x^3 + 6*E*x^3))*Log[4 - E^x] + (-12*x^2 + 3
*E^x*x^2)*Log[4 - E^x]^2)/(-64 - 32*E*x - 4*E^2*x^2 + E^x*(16 + 8*E*x + E^2*x^2) + (-32 - 8*E*x + E^x*(8 + 2*E
*x))*Log[4 - E^x] + (-4 + E^x)*Log[4 - E^x]^2),x]

[Out]

x^3 + x^4/(4 + E*x + Log[4 - E^x])

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fricas [A]  time = 0.88, size = 42, normalized size = 1.83 \begin {gather*} \frac {x^{4} e + x^{4} + x^{3} \log \left (-e^{x} + 4\right ) + 4 \, x^{3}}{x e + \log \left (-e^{x} + 4\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24*x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*
x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-9
6*x^3)*exp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*x*exp(1)+8)*exp(x)-8*x*exp(1)-32)*log(-exp(x)+4
)+(x^2*exp(1)^2+8*x*exp(1)+16)*exp(x)-4*x^2*exp(1)^2-32*x*exp(1)-64),x, algorithm="fricas")

[Out]

(x^4*e + x^4 + x^3*log(-e^x + 4) + 4*x^3)/(x*e + log(-e^x + 4) + 4)

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giac [A]  time = 0.88, size = 42, normalized size = 1.83 \begin {gather*} \frac {x^{4} e + x^{4} + x^{3} \log \left (-e^{x} + 4\right ) + 4 \, x^{3}}{x e + \log \left (-e^{x} + 4\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24*x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*
x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-9
6*x^3)*exp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*x*exp(1)+8)*exp(x)-8*x*exp(1)-32)*log(-exp(x)+4
)+(x^2*exp(1)^2+8*x*exp(1)+16)*exp(x)-4*x^2*exp(1)^2-32*x*exp(1)-64),x, algorithm="giac")

[Out]

(x^4*e + x^4 + x^3*log(-e^x + 4) + 4*x^3)/(x*e + log(-e^x + 4) + 4)

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maple [A]  time = 0.11, size = 24, normalized size = 1.04

method	result	size
risch	\(\frac {x^{4}}{\ln \left (-{\mathrm e}^{x}+4\right )+4+x \,{\mathrm e}}+x^{3}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*exp(x)*x^2-12*x^2)*ln(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24*x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*x^2)*ln
(-exp(x)+4)+(3*x^4*exp(1)^2+(3*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-96*x^3)*e
xp(1)-64*x^3-192*x^2)/((exp(x)-4)*ln(-exp(x)+4)^2+((2*x*exp(1)+8)*exp(x)-8*x*exp(1)-32)*ln(-exp(x)+4)+(x^2*exp
(1)^2+8*x*exp(1)+16)*exp(x)-4*x^2*exp(1)^2-32*x*exp(1)-64),x,method=_RETURNVERBOSE)

[Out]

x^4/(ln(-exp(x)+4)+4+x*exp(1))+x^3

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maxima [A]  time = 0.55, size = 41, normalized size = 1.78 \begin {gather*} \frac {x^{4} {\left (e + 1\right )} + x^{3} \log \left (-e^{x} + 4\right ) + 4 \, x^{3}}{x e + \log \left (-e^{x} + 4\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24*x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*
x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-9
6*x^3)*exp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*x*exp(1)+8)*exp(x)-8*x*exp(1)-32)*log(-exp(x)+4
)+(x^2*exp(1)^2+8*x*exp(1)+16)*exp(x)-4*x^2*exp(1)^2-32*x*exp(1)-64),x, algorithm="maxima")

[Out]

(x^4*(e + 1) + x^3*log(-e^x + 4) + 4*x^3)/(x*e + log(-e^x + 4) + 4)

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mupad [B]  time = 0.98, size = 139, normalized size = 6.04 \begin {gather*} \frac {\frac {x^2\,\left (64\,x+x^2\,{\mathrm {e}}^x-3\,x^2\,{\mathrm {e}}^{x+1}+12\,x^2\,\mathrm {e}-16\,x\,{\mathrm {e}}^x\right )}{{\mathrm {e}}^{x+1}-4\,\mathrm {e}+{\mathrm {e}}^x}-\frac {4\,x^3\,\ln \left (4-{\mathrm {e}}^x\right )\,\left ({\mathrm {e}}^x-4\right )}{{\mathrm {e}}^{x+1}-4\,\mathrm {e}+{\mathrm {e}}^x}}{\ln \left (4-{\mathrm {e}}^x\right )+x\,\mathrm {e}+4}+\frac {x^3\,\left (3\,\mathrm {e}+15\right )}{3\,\left (\mathrm {e}+1\right )}-\frac {16\,x^3}{\left ({\mathrm {e}}^x-\frac {4\,\mathrm {e}}{\mathrm {e}+1}\right )\,{\left (\mathrm {e}+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4 - exp(x))*(24*x^3*exp(1) - exp(x)*(6*x^3*exp(1) + 24*x^2 + 4*x^3) + 96*x^2 + 16*x^3) - log(4 - exp(
x))^2*(3*x^2*exp(x) - 12*x^2) - exp(x)*(exp(1)*(24*x^3 + 3*x^4) + 3*x^4*exp(2) + 48*x^2 + 16*x^3 - x^4) + exp(
1)*(96*x^3 + 12*x^4) + 12*x^4*exp(2) + 192*x^2 + 64*x^3)/(32*x*exp(1) - log(4 - exp(x))^2*(exp(x) - 4) - exp(x
)*(8*x*exp(1) + x^2*exp(2) + 16) + 4*x^2*exp(2) + log(4 - exp(x))*(8*x*exp(1) - exp(x)*(2*x*exp(1) + 8) + 32)
+ 64),x)

[Out]

((x^2*(64*x + x^2*exp(x) - 3*x^2*exp(x + 1) + 12*x^2*exp(1) - 16*x*exp(x)))/(exp(x + 1) - 4*exp(1) + exp(x)) -
(4*x^3*log(4 - exp(x))*(exp(x) - 4))/(exp(x + 1) - 4*exp(1) + exp(x)))/(log(4 - exp(x)) + x*exp(1) + 4) + (x^
3*(3*exp(1) + 15))/(3*(exp(1) + 1)) - (16*x^3)/((exp(x) - (4*exp(1))/(exp(1) + 1))*(exp(1) + 1)^2)

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sympy [A]  time = 0.25, size = 19, normalized size = 0.83 \begin {gather*} \frac {x^{4}}{e x + \log {\left (4 - e^{x} \right )} + 4} + x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x**2-12*x**2)*ln(-exp(x)+4)**2+((6*x**3*exp(1)+4*x**3+24*x**2)*exp(x)-24*x**3*exp(1)-16*x
**3-96*x**2)*ln(-exp(x)+4)+(3*x**4*exp(1)**2+(3*x**4+24*x**3)*exp(1)-x**4+16*x**3+48*x**2)*exp(x)-12*x**4*exp(
1)**2+(-12*x**4-96*x**3)*exp(1)-64*x**3-192*x**2)/((exp(x)-4)*ln(-exp(x)+4)**2+((2*x*exp(1)+8)*exp(x)-8*x*exp(
1)-32)*ln(-exp(x)+4)+(x**2*exp(1)**2+8*x*exp(1)+16)*exp(x)-4*x**2*exp(1)**2-32*x*exp(1)-64),x)

[Out]

x**4/(E*x + log(4 - exp(x)) + 4) + x**3

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